Approximaths

Regular Problem II

Illustration of another regular problem:

$\displaystyle x^2-2x -1 = 0 \quad \text{Problem}$

$\displaystyle x^2-2\epsilon x-1=0 \quad \text{Insert }\epsilon$

The solutions to the perturbed problem are (see figure below):

$\displaystyle x_1(\epsilon) = \frac{2\epsilon + \sqrt{4 \epsilon^2 + 4}}{2} = \epsilon + \sqrt{1 + \epsilon^2}$

$\displaystyle x_2(\epsilon) = \frac{2\epsilon - \sqrt{4 \epsilon^2 + 4}}{2} = \epsilon - \sqrt{1 + \epsilon^2}$

The figure above shows solutions of the problem $x^2-2\epsilon x-1 = 0$ using perturbation theory. For $\epsilon =0$ the solutions are -1 and 1 and for $\epsilon =1$ the solutions are exactly $1+\sqrt{2}$ and $1-\sqrt{2}$ .

$\displaystyle x_1(\epsilon) = 1 + \epsilon + \frac{1}{2}\epsilon^2 - \frac{1}{8}\epsilon^4 + \frac{1}{16}\epsilon^6 - \frac{5}{128}\epsilon^8 + O(\epsilon^{10})$

$\displaystyle x_2(\epsilon) = -1 + \epsilon - \frac{1}{2}\epsilon^2 + \frac{1}{8}\epsilon^4 - \frac{1}{16}\epsilon^6 + \frac{5}{128}\epsilon^8 + O(\epsilon^{10})$

Setting $\epsilon = 1$ :

$\displaystyle x_1(1) \approx 2 + \frac{1}{2} - \frac{1}{8} + \frac{1}{16} - \frac{5}{128} = 2.3984375$

$\displaystyle x_2(1) \approx 0 - \frac{1}{2} + \frac{1}{8} - \frac{1}{16} + \frac{5}{128} = -0.3984375$

This simple example clearly illustrates the application of regular perturbation theory to a quadratic equation, showing how the solutions vary continuously with the perturbation parameter $\epsilon$ .

May 28, 2026
Regular Problem I
In perturbation theory we distinguish regular and singular problems. In this section we aim to solve the original regular problem:

$\displaystyle x^2 - 2 = 0 \quad \text{Original problem}$

$\displaystyle x^2 - (1 + \epsilon) = 0 \quad \text{Insert epsilon}$

The solutions are:

$\displaystyle x_1(\epsilon) = -\sqrt{1+\epsilon}$

$\displaystyle x_2(\epsilon) = \sqrt{1+\epsilon}$

At $\epsilon=0$ the problem is solvable analytically:

$\displaystyle x_1 = -1, \quad x_2 = 1$

At $\epsilon=1$ we recover the solution of the original problem:

$\displaystyle x_1 = -\sqrt{2}, \quad x_2 = \sqrt{2}$

Figure. Regular problem: $x^2-(1+\epsilon)=0$ connecting $x^2-1=0$ to $x^2-2=0$ .

Characteristics of a regular perturbation problem:
- The solution depends smoothly on ε.
- The deformation from ε=0 to ε=1 is continuous.
- The problem at ε=0 is simple and solvable analytically.
- The original problem is recovered exactly at ε=1.
- No singularities or discontinuities appear.
- The solution admits a regular power-series expansion in ε.
We treat $\epsilon$ as a parameter to track the perturbation. The solution $x_2(\epsilon) = \sqrt{1+\epsilon}$ can be expressed as a Taylor series around the simple problem ( $\epsilon = 0$ ):

$\displaystyle x_2(\epsilon) = 1 + \frac{1}{2}\epsilon - \frac{1}{8}\epsilon^2 + \frac{1}{16}\epsilon^3 - \frac{5}{128}\epsilon^4 + \dots$

The original problem $x^2 - 2 = 0$ is recovered by evaluating this series at $\epsilon = 1$ . The smooth dependence on the parameter allows us to reach the target solution. For $\epsilon=1$ , the first few terms give:

$\displaystyle x_2(1) \approx 1 + 0.5 - 0.125 + 0.0625 - 0.039 = 1.3985$

which is already a good approximation $\sqrt{2} \approx 1.4142$ .

This example illustrates a regular homotopy connecting the simple problem $x^2 - 1 = 0$ to the target problem $x^2 - 2 = 0$ .
May 21, 2026

Polynomial expansion

Expanding series and sorting coefficients by hand can be tedious. The Python library sympy allows us to automate this process efficiently.

Polynomial expansion and coefficient extraction

from sympy import symbols, expand, collect

# define variables
a0, a1, a2, a3, x = symbols('a0 a1 a2 a3 x')

# expand the expression
expr = expand((a0 + a1*x + a2*x**2 + a3*x**3)**2)

print("Expanded expression:")
print(expr)

print("\nCollected by powers of x:")
print(collect(expr, x))

Output:

Expanded expression:
a0**2 + 2*a0*a1*x + 2*a0*a2*x**2 + 2*a0*a3*x**3
+ a1**2*x**2 + 2*a1*a2*x**3 + 2*a1*a3*x**4
+ a2**2*x**4 + 2*a2*a3*x**5 + a3**2*x**6

Collected by powers of x:
a0**2 + 2*a0*a1*x
+ (2*a0*a2 + a1**2)*x**2
+ (2*a0*a3 + 2*a1*a2)*x**3
+ (2*a1*a3 + a2**2)*x**4
+ 2*a2*a3*x**5
+ a3**2*x**6

We can also extract coefficients order by order:

coeffs = [expr.coeff(x, i) for i in range(7)]

for i, c in enumerate(coeffs):
    print(f"Coefficient of x^{i}:", c)

Output:

Coefficient of x^0: a0**2
Coefficient of x^1: 2*a0*a1
Coefficient of x^2: 2*a0*a2 + a1**2
Coefficient of x^3: 2*a0*a3 + 2*a1*a2
Coefficient of x^4: 2*a1*a3 + a2**2
Coefficient of x^5: 2*a2*a3
Coefficient of x^6: a3**2

This approach allows us to systematically extract and match coefficients order by order, which is particularly useful in perturbation theory.

May 14, 2026

sqrt(2) bis

Let us revisit the problem of computing the value of $\sqrt{2}$ , but this time we slightly modify our perturbative approach to improve convergence.

Previously, we expanded around $x=1$ , which is relatively far from the true value. Instead, we now choose a better starting point.

$\displaystyle x = \sqrt{2} \quad \text{Problem}$

We observe that:

$\displaystyle \left(\frac{3}{2}\right)^2 = \frac{9}{4} = 2.25 \quad \text{so it is close to } 2$

This suggests writing:

$\displaystyle x = \frac{3}{2} + \delta$

The correction $\delta$ is expected to be small, which improves the convergence of the perturbative expansion.

We now insert this into the equation:

$\displaystyle x^2 = 2$

$\displaystyle \left(\frac{3}{2} + \delta \right)^2 = 2$

$\displaystyle \frac{9}{4} + 3\delta + \delta^2 = 2$

Rearranging, we obtain:

$\displaystyle 3\delta + \delta^2 = -\frac{1}{4}$

We now introduce a perturbation parameter $\epsilon$ :

$\displaystyle 3\delta + \delta^2 = -\frac{1}{4}\varepsilon$

We seek a solution of the form:

$\displaystyle \delta = b_1 \varepsilon + b_2 \varepsilon^2 + b_3 \varepsilon^3 + \cdots$

Substituting into the equation and expanding:

$\displaystyle 3b_1\varepsilon + 3b_2\varepsilon^2 + b_1^2\varepsilon^2 + \cdots = -\frac{1}{4}\varepsilon$

Matching powers of $\varepsilon$ :

$\displaystyle \varepsilon^1: \quad 3b_1 = -\frac{1}{4} \Rightarrow b_1 = -\frac{1}{12}$

$\displaystyle \varepsilon^2: \quad 3b_2 + b_1^2 = 0 \Rightarrow b_2 = -\frac{1}{432}$

Thus:

$\displaystyle \delta = -\frac{1}{12}\varepsilon - \frac{1}{432}\varepsilon^2 + \cdots$

We finally obtain:

$\displaystyle x = \frac{3}{2} - \frac{1}{12}\varepsilon - \frac{1}{432}\varepsilon^2 + \cdots$

Setting $\varepsilon = 1$ :

$\displaystyle x \approx \frac{3}{2} - \frac{1}{12} - \frac{1}{432} = 1.41435$

which is already a very good approximation of $\sqrt{2} \approx 1.41421$ .

This example illustrates a key idea in perturbation theory:

The choice of the unperturbed problem strongly affects the convergence of the series.

By expanding around a value closer to the true solution, we obtain a much more efficient approximation with fewer terms.

May 7, 2026
sqrt(2)

Let’s imagine we don’t have a calculating machine or computer and we want to calculate the value of the root of 2 using perturbation theory. Remember that we always insert epsilon so that the problem is exactly solvable for epsilon equals zero. Here’s how we might proceed:

$\displaystyle x = \sqrt{2} \quad \text{Problem}$

$\displaystyle x^2 = 2$

$\displaystyle x^2 - 1 = 1$

$\displaystyle x^2 - \epsilon = 1 \quad \text{Insert }\epsilon$

$\displaystyle (a_0 + a_1 \epsilon + a_2 \epsilon^2 + ...)^2 - \epsilon = 1 \quad \text{Insert }x_{\epsilon}$

$\displaystyle a_0^2 + 2a_0 a_1 \epsilon + 2 a_0 a_2 \epsilon^2 + a_1^2 \epsilon^2 + ... - \epsilon = 1$

$\displaystyle a_0^2 + (2a_0 a_1 - 1) \epsilon + (2 a_0 a_2 + a_1^2) \epsilon^2 + ... = 1$

In this case, we keep only the powers of epsilon up to order two. We want to perform a perturbative calculation of order two.

$\displaystyle \implies a_0 = 1, a_1 = \frac{1}{2}, a_2 = -\frac{1}{8}, ...$

$\displaystyle x_{\epsilon} = 1 + \frac{1}{2} \epsilon - \frac{1}{8}\epsilon^2 + ...$

$\displaystyle x_{\epsilon = 1} = 1 + \frac{1}{2} - \frac{1}{8} + ...$

Let us solve the same problem but this time we keep the powers of epsilon up to order three.

$\displaystyle x = \sqrt{2} \quad \text{Problem}$

$\displaystyle x^2 = 2$

$\displaystyle x^2 - 1 = 1$

$\displaystyle x^2 - \epsilon = 1 \quad \text{Insert }\epsilon$

$\displaystyle (a_0 + a_1 \epsilon + a_2 \epsilon^2 + a_3 \epsilon^3 + ...)^2 - \epsilon = 1 \quad \text{Insert }x_{\epsilon}$

$\displaystyle a_0^2+a_1^2\epsilon^2+2a_0a_1\epsilon+2a_0a_2\epsilon^2+2a_0a_3\epsilon^3+2a_1a_2\epsilon^3 + ... - \epsilon = 1$

$\displaystyle a_0^2+(2a_0a_1 -1)\epsilon+(2a_0a_2 + a_1^2)\epsilon^2+(2a_0a_3+2a_1a_2)\epsilon^3 + ... = 1$

$\displaystyle \implies a_0 = 1, a_1 = \frac{1}{2}, a_2 = -\frac{1}{8}, a_3 = \frac{1}{16}$

$\displaystyle x_{\epsilon} = 1 + \frac{1}{2} \epsilon - \frac{1}{8}\epsilon^2 + \frac{1}{16}\epsilon^3 + ...$

$\displaystyle x_{\epsilon = 1} = 1 + \frac{1}{2} - \frac{1}{8} + \frac{1}{16}$

This suggests the following (using the double factorial notation):

$\displaystyle \sqrt{2} = 1 + \sum_{k=1}^\infty (-1)^{k+1} \frac{(2k-3)!!}{(2k)!!}$

$\displaystyle = 1 + \frac{1}{2} - \frac{1}{2\cdot4} + \frac{1\cdot3}{2\cdot4\cdot6} - \frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} + \cdots$

$\displaystyle = 1 + \frac{1}{2} - \frac{1}{8} + \frac{1}{16} - \frac{5}{128} + \frac{7}{256} + \cdots$

We could use Euler’s summation to speed up the convergence of this series. We observe that perturbing the initial problem in this form:

$\displaystyle x^2 - \epsilon = 1$

is similar to finding the roots of the polynomial

$\displaystyle x^2 - \epsilon -1 = 0$

$\displaystyle x^2 - (\epsilon +1) = 0$

whose solution is (see figure below)

$\displaystyle x= \sqrt{\epsilon+1}$

April 30, 2026
Perturbation Theory – Case Study 2

We would like to solve a very simple problem. This time we will set a power of $\epsilon$ :

$\displaystyle x - 2 = 0 \quad \text{Problem}$

$\displaystyle x - 2\epsilon^2 = 0 \quad \text{Insert }\epsilon^2$

$\displaystyle x = 0 \quad \text{Set }\epsilon^2 = 0$

$\displaystyle (a_0 + a_1 \epsilon + a_2 \epsilon^2 + \dots) - 2\epsilon^2 = 0 \quad \text{Insert }x_{\epsilon}$

$\displaystyle a_0 + a_1 \epsilon + (a_2 - 2)\epsilon^2 + \dots = 0 + 0 + 0 + \dots$

By identifying coefficients of equal powers of $\epsilon$ (uniqueness of power series expansion), we obtain:

$\displaystyle \implies a_0 = 0,\ a_1 = 0,\ a_2 = 2,\ \dots$

$\displaystyle x_{\epsilon} = 0 + 0 + 2\epsilon^2 + \dots$

$\displaystyle x_{\epsilon} = 2\epsilon^2$

Evaluating the solution at $\epsilon = 1$ recovers the original problem and gives $x = 2$ . In the figure below we visualize the solution of $x - 2\epsilon^2 = 0$ as a function of $\epsilon$ .

For this very simple example, we essentially chose a placement of $\epsilon$ such that the unperturbed problem (for $\epsilon = 0$ ) becomes trivial. We also have many choices for the power of $\epsilon$ . In this example, the perturbation series terminates and yields the exact solution.

April 23, 2026
Perturbation Theory – Case Study 1
Since then, we have presented some methods to sum series (Padé approximants, Euler summation, Borel summation, Borel-Écalle summation, generic summation and Zeta summation). We have also seen techniques to accelerate convergence (Shanks transformation and Richardson transformation).

When a problem is not solvable exactly (extracting the roots of a polynomial, solving a specific differential equation, …) the idea of perturbation theory is to obtain an approximate solution (analytically) by inserting a small parameter $\epsilon$ and assuming the answer has the form $a_0 + a_1 \epsilon + a_2 \epsilon^2 + a_3 \epsilon^3 + \dots$ .

Procedure:
- Insert $\displaystyle \epsilon$
- Set $\displaystyle \epsilon = 0$ (the problem must be exactly solvable for $\displaystyle \epsilon = 0$ )
- The solution has the form $\displaystyle x_{\epsilon} = a_0 + a_1 \epsilon + a_2 \epsilon^2 + a_3 \epsilon^3 + \dots$
- Set $\displaystyle \epsilon = 1$ (the solution of the initial problem is now $\displaystyle a_0 + a_1 + a_2 + a_3 + \dots$ )
- Study the convergence of $\displaystyle a_0 + a_1 + a_2 + a_3 + \dots$
- If necessary use a summation method
Solve a very simple ‘problem’ using perturbation theory:

$\displaystyle x - 2 = 0 \quad \text{Problem}$

$\displaystyle x - 2\epsilon = 0 \quad \text{Insert } \epsilon$

$\displaystyle (a_0 + a_1 \epsilon + a_2 \epsilon^2 + \dots) - 2\epsilon = 0 \quad \text{Insert } x_{\epsilon}$

$\displaystyle a_0 + (a_1 - 2)\epsilon + a_2 \epsilon^2 + \dots = 0 + 0 + 0 + \dots$

Because coefficients of Taylor series are unique (see proof) we are allowed to compare powers of $\epsilon$ .

$\displaystyle \implies a_0 = 0,\ a_1 = 2,\ a_2 = 0,\ \dots$

$\displaystyle x_{\epsilon} = 0 + 2\epsilon + 0 + \dots$

$\displaystyle x_{\epsilon} = 2\epsilon$

This implies that $x = 2$ (setting $\epsilon = 1$ ).

In the figure below, we visualize the solution of $x - 2\epsilon = 0$ as a function of $\epsilon$ .

Solution of the problem $x - 2 = 0$ using perturbation theory. For $\epsilon = 0$ the solution is zero and for $\epsilon = 1$ the solution is 2. By making $\epsilon$ approach 1 we obtain an increasingly precise answer to the initial problem.
April 16, 2026
The Path to Perturbation Theory

“Almost no problems are exactly solvable”

Carl Bender

In physics, most problems—from the anharmonic oscillator to Quantum Electrodynamics (QED)— cannot be solved exactly. We rely on Perturbation Theory, expanding the solution in powers of a small coupling constant (often denoted λ or α).

A classic example is the quantum anharmonic oscillator, where the Hamiltonian is

$\displaystyle H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2 + \lambda x^4$

(with the quartic term treated as a small perturbation λ ≪ 1). Another famous case is QED, where the fine-structure constant α ≈ 1/137 serves as the small parameter for expanding processes involving electrons and photons.

The challenge? These perturbative expansions are almost always asymptotic series with a null radius of convergence. The coefficients $a_n$ typically grow factorially as $n!$ (or even faster), making the series eventually diverge regardless of how small the coupling constant is.

For instance, in the anharmonic oscillator, the perturbative corrections to the energy levels involve terms whose coefficients grow like $n!$ , leading to a divergent series for any nonzero λ. Similarly, in QED, the perturbative series for quantities like the electron’s anomalous magnetic moment (g−2) is asymptotic: it gives spectacular agreement with experiment up to high orders, but diverges if pushed too far.

This is where our journey pays off. By applying Borel resummation and Padé approximants to these divergent series, we can extract non-perturbative information and obtain remarkably accurate physical predictions. In quantum field theory, Borel summation helps recover meaningful results from perturbative expansions (for example, in certain Euclidean field theories or by handling contributions from instantons and renormalons). Padé approximants are routinely used to accelerate convergence and estimate the behavior of series in QCD and other models.

In our upcoming posts, we will see how this mathematical framework becomes the essential language of modern theoretical physics.

April 9, 2026
Summation Techniques for Perturbation Theory
After several months of exploring various summation and acceleration methods, it is time to synthesize these tools before we apply them to one of their most important domains: Perturbation Theory.

When a power series $\displaystyle \sum_{n=0}^\infty a_n z^{n}$ has a small or even zero radius of convergence, or converges too slowly to be numerically useful, we require more than standard arithmetic. Here is a recap of the methods we have covered and how they bridge the gap between formal series and numerical values.

Summation Methods: Taming Divergence
- Borel and Borel–Écalle: By using the Borel transform, we transform factorially growing coefficients into a convergent series in the Borel plane. The Borel–Écalle framework further allows us to handle singularities via resurgence theory, providing a unique “resummed” value even for non-Borel-summable series.
- Euler and Zeta summation: These methods assign finite values to divergent sums by analytic continuation. While Euler summation is ideal for alternating series, Zeta summation provides a powerful way to handle the infinite sums that frequently appear in quantum vacuum energy calculations.
- Generic summation: Beyond these classical approaches, we have also explored the idea of generic summation, which provides a unifying framework for assigning values to divergent or slowly convergent series.
Acceleration Techniques: Enhancing Convergence
- Shanks Transformation and Padé Approximants: These nonlinear transformations (often related to continued fractions) excel at capturing the behavior of functions beyond their radius of convergence, particularly when poles are present.
- Richardson Extrapolation: A fundamental tool for numerical analysis that cancels out the leading error terms, allowing us to estimate the limit of a sequence S_n with much higher precision from only a few terms.
An important consistency check underlies all these techniques: whenever different summation or acceleration methods apply to the same series, they agree on a common value. This convergence of independent approaches is not accidental, it reflects the fact that these methods capture an underlying analytic object beyond the formal series itself. When they work, they do not merely assign a value: they reveal a coherent extension of the function that the original divergent expansion was hinting at.
April 2, 2026
Richardson extrapolation: example

Richardson extrapolation acts as a powerful convergence accelerator in numerical methods. Instead of relying solely on extremely fine discretizations—which can be computationally expensive—it combines results from coarser simulations to estimate the zero-step (continuum) limit.

This idea appears in several numerical techniques (e.g., Romberg integration, mesh convergence studies in engineering), where it can significantly improve accuracy at a modest additional cost.

When the error behaves regularly, Richardson extrapolation can dramatically reduce the error—sometimes by an order of magnitude or more—without requiring prohibitively fine meshes.

Consider the following (slowly) converging series (see this post Zeta summation: Basel problem):

$\displaystyle 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ... = \frac{\pi^2}{6}$

and calculate its corresponding Richardson extrapolation using up to 4-term partial sums:

$\displaystyle N = 2$

$\displaystyle S_2 = 1 + \frac{1}{4} = \frac{5}{4}$

$\displaystyle S_3 = 1 + \frac{1}{4} + \frac{1}{9} = \frac{49}{36}$

$\displaystyle S_4 = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} = \frac{205}{144}$

The Richardson extrapolation is (see this post Richardson extrapolation):

$\displaystyle \mathcal{R}_N = \frac{N^2S_N -2(N+1)^2S_{N+1} + (N+2)^2S_{N+2}}{2}$

in this case:

$\displaystyle \mathcal{R}_2 = \frac{4 \times (\frac{5}{4}) - 2 \times 9 \times (\frac{49}{36}) + 16 \times \frac{205}{144}}{2}$

$\displaystyle = \frac{5 - \frac{49}{2} + \frac{205}{9}}{2}$

$\displaystyle = 1.6389$

We have:

$\displaystyle \frac{\pi^2}{6} = 1.6449$

$\displaystyle \mathcal{R}_{2} = 1.6389$

$\displaystyle S_4 = 1.4236$

Using solely a partial series with 4 terms, we were able to speed up the convergence of the above-mentioned series considerably using Richardson extrapolation.

March 26, 2026